Solve Right Triangle Given Perimeter and Altitude  Problem With Solution
Solve a right triangle given its perimeter and altitude. A problem with detailed solution.
Problem :
The length of the perimeter of a right triangle is equal to 240 cm. The length of its altitude perpendicular to its hypotenuse is equal to 48 cm. Find the two sides and the hypotenuse of the triangle.
Solution to Problem :

Using the figure below, we can write an equation for the perimeter and a second equation for the altitude as follows
240 = a + b + H and h = 48

Pythagora's theorem applied to the above right triangle gives
a^{ 2} + b^{ 2} = H^{ 2}

The area A of the right triangle may be calculated using the sides a and b as follows
A = (1 / 2) a b

The same area A may be calculated using the altitude and base (hypotenuse) as follows
A = (1 / 2) h H

The above formulas for the same area A gives the equation
a b = h H

Since h = 48, the above becomes
a b = 48 H

We now use the perimeter equation 240 = a + b + H to write
a + b = 240  H

Square both sides of the equation to obtain
(a + b)^{ 2} = (240  H)^{ 2}
a^{ 2} + b^{ 2} + 2 a b = 240^{ 2} + H^{ 2}  480 H

We now subtract the equation a^{ 2} + b^{ 2} = H^{ 2} from the above equation to obtain
2 a b = 240^{ 2}  480 H

We now use the fact that a b = 48 H to write the above equation as
2 (48 H) = 240^{ 2}  480 H

Solve the above equation for H to obtain
H = 100 cm

We now substitute H by 100 in the two equations
a + b = 240  H and a b = 48 H to obtain
a + b = 140 and a b = 4800

We now use the equation a b = 4800 to write
b = 4800 / a

We now substitute b in a + b = 140 by 4800 / a to write an equation in a only
a + 4800 / a = 140

Multiply all terms of the above equation to obtain a quadratic equation
a^{ 2} + 4800 = 140 a

Solve the above for a to obtain two solutions
a = 80 cm and a = 60 cm

Use the equation b = 4800 / a to find b
for a = 80 cm then b = 60 cm
and
for a = 60 cm b = 80 cm.

If we assume that a < b, the solution to our problem is
a = 60 cm, b = 80 cm and H = 100 cm.
More references to geometry problems.
Geometry Tutorials, Problems and Interactive Applets.

